So I was experimenting with half-blocks and stairs, and made this structure:

half-block and stair prison sketch half-block and stair prison screenshot

and since I was inside it when I placed the last stair, I found that I was trapped.

This surprised me since the space inside the structure is 2 blocks tall and 2 blocks wide, and the gap between the stairs in the corner looks quite large. Doing some basic geometry, it's clear that the gap is 1/sqrt(2) or a little more than 0.7 units wide (where 1 unit is the width of one side of a block). It is already obvious that Steve can fit through a 1 unit wide space, but noticing also that doors have width(thickness) and Steve can still fit through the space in an open door seems to imply that Steve's width is somewhat less than 1 unit but more than 0.7 units. I've checked out the Minecraft Wiki and did not find anything detailing this aspect.

How fat is Steve (and the other Steve-sized mobs)?

Edit: I did a bit more research and found that doors take up 3/16 pixels of 1 unit, leaving 13/16 or 0.8125 units of space that Steve can fit through, with some wiggle room left. This can be seen in the following screenshot:

grass pixels with door in block

So the range now is 0.7071 < Steve < 0.8125.

  • 10
    Steve isn't fat, s/he's pleasantly plump!
    – user3389
    Aug 9, 2012 at 6:02
  • 1
    Steve is 1.80 meters tall (30 pixels) :D Aug 9, 2012 at 9:44
  • 29
    :( downvoting this question
    – Steve V.
    Aug 9, 2012 at 14:28
  • 3
    @Steve ... you weren't supposed to find out about this ... I'm sorry to tell you this way, but you're bigger than 0.7 units, and not in the middle either, the middle is open. Ps. Image of structure still to come in future edit (later today).
    – Dani
    Aug 9, 2012 at 14:57
  • 6
    Steve isn't fat, he's just big-voxeled.
    – gnovice
    Aug 9, 2012 at 18:44

3 Answers 3


Put Steve in a 1x1 hole and measure the wiggle room. ALL the room - WiggleRoom = Steve

In a 1x1 hole, I pointed the camera straight down and moved from corner to corner. Regardless of corner or orientation, the crosshairs pointed at the farthest point of a 5x5 square from the corner.

16 = 5 + 6 + 5

That's 16 total square width, 5 for the each side of Steve, and 6 for the wiggle room.

Front/back/left/right, the closest you can get to a wall is 5. This test is unable to confirm if Steve is 10x10 square, or a 5 radius circle.

Note: this finding contradicts the original question, as Steve is 0.625 wide, and the asker was expecting more than that. However the original question also supports this answer in the door screenshot (5 texture pixels from door).

It seems reasonable to think that Steve is an unrotating 0.625x0.625 square, which is 0.88 from corner to corner, and could not fit through the ~0.7 gap.

Here is Steve (in green), blocked by two stairs (in purple). Steve is 10 units big, but the yellow line is ~14.14 units big, which is too large to go through the ~11.3 unit opening.

Steve trapped

  • 6
    Not so. There is plenty of wiggle room. You should have done some research before posting. Also, doors plainly and clearly have thickness as is provable by running into the side of an open door with no walls; or by standing in the door's 1x1 block space with blocks all around and counting the pixels of the floor, or comparing wiggle room in both directions of the floor's plane.
    – Dani
    Aug 9, 2012 at 23:07
  • I think @AmyB was saying that if you put him in the whole you can tell his less then 1 by 1 as there is still Wiggle Room. E.G Steve is as big as 1 by 1 minus how much he can wiggle around.
    – TrewTzu
    Aug 10, 2012 at 5:44
  • 1
    @TrewTzu Yes, but he edited his answer completely from the original text, which is why my comment does not make any sense, but I think it should remain anyways, since you can view the original answer by clicking on the edited date. Also, I've tried this and there is no accurate way of measuring "wiggle room".
    – Dani
    Aug 13, 2012 at 2:09
  • You can't stand facing directly forward and move from side to side, checking your position at each extreme?
    – Random832
    Aug 13, 2012 at 20:02
  • @Random832 I've tried this, with the F3 coordinates, but I wasn't able to get an accurate measure. I also tried comparing the crosshair movement, but this is dependent on the distance from the block in front of you, so it was even more inaccurate. If you are able to take a measurement using this or another technique, please post as an answer with your proof.
    – Dani
    Aug 14, 2012 at 16:09
  • Torso is half a meter wide.
  • Shoulders at furthest point are a meter apart
  • Legs are a quarter of a meter wide each

From here

Your legs are 3/4 meters long, as are your arms and your body's height. Your head is half a meter. You are not an accurately proportioned human, but you are realistically tall for a modern human.

EDIT: Your torso is half a meter wide. Your shoulders at furthest point are a meter apart. Your legs are a quarter of a meter wide each. YOU ARE FAT IN MINECRAFT.

  • 8
    And the Body Mass Index?
    – fredley
    Aug 9, 2012 at 8:33
  • 2
    you'll get a divide by zero error Aug 9, 2012 at 8:47
  • 4
    This doesn't seem to be the correct answer, at least for the characters hitbox. This answer says that in total, you should be one meter wide, but as the question states, open doors have a width, so that can't possibly be true or you wouldn't be able to walk through them. It's probably correct based entirely on the player model though.
    – Mr Smooth
    Aug 9, 2012 at 9:36
  • 6
    "Your head is half a meter." O_O
    – Alex
    Aug 9, 2012 at 9:50
  • @Alex half a meter tall that is Aug 13, 2012 at 7:24

Can't we just use Steve himself to know his body size?

I will be using Steve's pixels (sp) as unit to measure.

Steve's head is 8 sp tall, his torso is 12sp and his leg is 12sp tall then he must be 32 sp tall. If Steve is 180cm tall then all we gotta do is divide 180 by 32, the result being 5.625cm per sp.

Head Steve's head is 8spx8spx8sp if we multiply that by 5.625 we get 45cmx45cmx45cm which is 91125cm^3

Torso Steve's torso is 12spx8spx4sp if we multiply that by 5.625 we get 67.7cmx45x22.5cm so his torso is 68546.25cm^3.

Arms Each arm is 12x4x4 multiply it by 5.625 and we get 67.7cmx22.5cmx22.5cm which is 34273.125cm^3

Legs The same as the arms. 12x4x4 is 67.7cmx22.5cmx22.5cm giving 34273.125cm^3 as result.

This way we can easily get the exact size of Steve so I don't belive that should be a problem to know how fat Steve is. I belive the difficlt part is defining "fat" for Steve. Accodring to the google dictionary "fat" is defined by "having a large amount of excess flesh." If we take Steve's direct body size he is definetly fat for a human but is Steve a human? I don't remember any time Steve has been described directly as a human by Mojang and considering the various abilities he shows throughot the game (like the carrying a stack of gold blocks without any problem) I don't think it would make sense to think of him as one. Steve is something else and thus we should considerate him as such. Calling Steve fat because if he were the same species as us he would it's the same as saying that an elephant is fat because if it were a human it would be. We need to know how "fat" in the Minecraft world is like to be able to know.

  • This is cosmetic Steve, and the OP seems to be worried about hitbox Steve, the Steve that can't escape the OP's structure. Nevertheless, good approach!
    – Joachim
    Nov 29 at 0:06

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